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3 years ago

If the pressure of a gas is 2.000 × 104 kPa, what is its pressure in atm?

Chemistry

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

197200atm

Explanation:

There are different units to express pressure of a gas, some of them are atmosphere, Bar, Kilopascal, Pascal, torr among other.

In this exercise we need to know the equivalence between atm and Kpa.

if, 1kp=0,00986atm

then,

(2,000 x104kPa)*(0,00986atm)/1Kpa= 0,00986 atm

luda_lava [24]3 years ago

4 0

Answer:

197.4 atm is the Answer.

Explanation:

Pressure can be expressed in unit of torr, mm Hg, psi, atm, kPa etc

1 atm = 760 torr = 760 mm Hg = 14.7 psi = 101.325 kPa

So Here to convert atm to kPa the conversion factor is either $\frac {1atm}{101.325kPa}$ or $\frac {101.325kPa}{1atm}$

We need the answer in atm so kPa should get cancel and it should be in the denominator

$2.000 \times 10^4 kPa\times \frac {1atm}{101.325kPa}$

=197.4 atm is the Answer.

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3 years ago

If you had 6 moles of CaCl2 and 5 moles of Na3PO4, which of these would be the limiting and excess reactant

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Answer:

Na3PO4 is excess reactant, CaCl2 is limiting reactant.

Explanation:

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3 years ago

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Answer:

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2 years ago

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3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

3 years ago

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